What is the tangent of y=4x^3-2x ?

The tangent of y=4x^3-2x is y=(12a_{0}^2-2)x-8a_{0}^3

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tangent of y=4x^3-2x

tangent of

y = 4 x^{3} - 2 x

y = (12 a_{0}^{2} - 2) x - 8 a_{0}^{3}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 4 a_{0}^{3} - 2 a_{0})

Find the slope of

y = 4 x^{3} - 2 x : \frac{d y}{d x} = 12 x^{2} - 2

Compute the slope of

y = 4 x^{3} - 2 x

x = a_{0} : m = 12 a_{0}^{2} - 2

Find the line with slope m=

12 a_{0}^{2} - 2

and passing through

(a_{0}, 4 a_{0}^{3} - 2 a_{0}) : y = (12 a_{0}^{2} - 2) x - 8 a_{0}^{3}

y = (12 a_{0}^{2} - 2) x - 8 a_{0}^{3}

x \to 5 lim (x (5))

\frac{d}{d x} (\frac{- 2}{x})

e^{2 x} + 2 x e^{2 x}

\frac{d}{d x} (cos (\frac{1 - e ^{5 x}}{1 + e ^{5 x}}))

\int \frac{1}{6 x ^{2}} d x

What is the tangent of y=4x^3-2x ?
The tangent of y=4x^3-2x is y=(12a_{0}^2-2)x-8a_{0}^3